find the area of the shaded region: x=y^2-2 and x=e^y, between the areas of y=1 and y=-1

To find the area of the shaded region bounded by the curves \(x = y^2 - 2\) and \(x = e^y\) between \(y = 1\) and \(y = -1\), we need to calculate the integral of the difference between these two functions over the given interval.

### Step-by-Step Solution:

1. **Find the points of intersection:**
   To find the points of intersection between the curves \(x = y^2 - 2\) and \(x = e^y\), we need to set them equal to each other:

   \[
   y^2 - 2 = e^y
   \]

   This equation does not have a simple algebraic solution, so let's check the values of \(y\) in the interval \([-1, 1]\):

   - For \(y = -1\):  
     \[
     (-1)^2 - 2 = 1 - 2 = -1, \quad e^{-1} \approx 0.3679
     \]

     The curves do not intersect here.

   - For \(y = 0\):  
     \[
     0^2 - 2 = -2, \quad e^0 = 1
     \]

     No intersection.

   - For \(y = 1\):  
     \[
     1^2 - 2 = -1, \quad e^1 \approx 2.718
     \]

     No intersection.

   Thus, the curves do not intersect between \(y = -1\) and \(y = 1\).

2. **Determine which function is on top:**
   We need to determine which function is greater on the interval \([-1, 1]\). Evaluating at points within this interval:

   - For \(y = -1\):  
     \(y^2 - 2 = -1\), \(e^y \approx 0.3679 \). Here, \(e^y > y^2 - 2\).

   - For \(y = 0\):  
     \(y^2 - 2 = -2\), \(e^y = 1 \). Again, \(e^y > y^2 - 2\).

   - For \(y = 1\):  
     \(y^2 - 2 = -1\), \(e^y \approx 2.718\). Again, \(e^y > y^2 - 2\).

   Therefore, \(e^y\) is the upper function and \(y^2 - 2\) is the lower function over this interval.

3. **Set up the integral to find the area:**
   The area \(A\) between the curves from \(y = -1\) to \(y = 1\) is given by:

   \[
   A = \int_{-1}^{1} \left(e^y - (y^2 - 2)\right) \, dy
   \]

   Simplify the integrand:

   \[
   A = \int_{-1}^{1} \left(e^y - y^2 + 2\right) \, dy
   \]

4. **Compute the integral:**

   The integral can be split into three parts:

   \[
   A = \int_{-1}^{1} e^y \, dy - \int_{-1}^{1} y^2 \, dy + \int_{-1}^{1} 2 \, dy
   \]

   Let's compute these integrals:

   - \(\int e^y \, dy = e^y\), so:

     \[
     \left[ e^y \right]_{-1}^{1} = e^1 - e^{-1} = e - \frac{1}{e}
     \]

   - \(\int y^2 \, dy = \frac{y^3}{3}\), so:

     \[
     \left[ \frac{y^3}{3} \right]_{-1}^{1} = \frac{1^3}{3} - \frac{(-1)^3}{3} = \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{2}{3}
     \]

   - \(\int 2 \, dy = 2y\), so:

     \[
     \left[ 2y \right]_{-1}^{1} = 2(1) - 2(-1) = 2 + 2 = 4
     \]

5. **Combine the results:**

   \[
   A = \left(e - \frac{1}{e}\right) - \frac{2}{3} + 4
   \]

   Simplify further:

   \[
   A = e - \frac{1}{e} + \frac{10}{3}
   \]

### Final Answer:

\[
\boxed{A = e - \frac{1}{e} + \frac{10}{3}}
\]

Would you like any further details or explanations?  

### Related Questions:
1. How would the area change if the interval was between \(y = 0\) and \(y = 2\)?
2. What is the volume of the solid formed when the shaded region is rotated about the x-axis?
3. What are the points of intersection between \(x = y^2 - 2\) and \(x = e^y\)?
4. How do you find the area between two curves if both are functions of \(x\)?
5. Can you find the area between the curves \(y = \ln(x)\) and \(y = x^2 - 4\) from \(x = 1\) to \(x = 3\)?

### Tip:
When finding the area between two curves, always check which function is on top within the given interval to ensure you subtract correctly!

This step-by-step solution demonstrates how to find the area of the shaded region between the curves x = y^2 - 2 and x = e^y over the interval y = -1 to y = 1. The problem involves calculating the integral of the difference between these functions, using key calculus concepts such as integration, exponential and quadratic functions, and definite integrals. Suitable for undergraduate calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation